An object is kept on a smooth inclined plane of height 1 unit and length l units. The horizontal acceleration to be imparted to the inclined plane so that the object is stationary relative to the incline is

Options

A. `gsqrt(l^2-1)`
B. `g(l^2 -1)`
C. `g/sqrt(l^2-1)`
D. `g/(l^2-1)`

Answer

To solve the problem, we need to determine the horizontal acceleration \( a \) that must be imparted to the inclined plane so that the object resting on it remains stationary relative to the incline.

### Step-by-Step Solution:

1. **Understand the Geometry of the Inclined Plane:**
- The height of the incline is given as \( h = 1 \) unit.
- The length of the incline is \( l \) units.
- We can find the angle of inclination \( \theta \) using the relationship:
\[ \sin \theta = \frac{\text{height}}{\text{hypotenuse}} = \frac{1}{l} \]
- Thus, the angle \( \theta \) can be expressed as:
\[ \theta = \arcsin\left(\frac{1}{l}\right) \]

2. **Identify Forces Acting on the Object:**
- The weight of the object \( mg \) acts vertically downwards.
- The normal force \( N \) acts perpendicular to the inclined surface.
- When the inclined plane accelerates horizontally with acceleration \( a \), a pseudo force \( ma \) acts on the object in the opposite direction of the acceleration.

3. **Resolve Forces Along the Inclined Plane:**
- The component of the gravitational force acting down the incline is:
\[ F_{\text{gravity}} = mg \sin \theta \]
- The component of the pseudo force acting up the incline is:
\[ F_{\text{pseudo}} = ma \cos \theta \]

4. **Set Up the Equation for Equilibrium:**
- For the object to remain stationary relative to the incline, the forces must balance:
\[ mg \sin \theta = ma \cos \theta \]

5. **Cancel Mass and Rearrange:**
- Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ g \sin \theta = a \cos \theta \]
- Rearranging gives:
\[ a = g \frac{\sin \theta}{\cos \theta} = g \tan \theta \]

6. **Find \( \tan \theta \):**
- From the right triangle formed by the incline, we know:
\[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{\sqrt{l^2 - 1}} \]
- Therefore, substituting this into our equation for \( a \):
\[ a = g \cdot \frac{1}{\sqrt{l^2 - 1}} \]

7. **Final Result:**
- The horizontal acceleration \( a \) that must be imparted to the inclined plane is:
\[ a = \frac{g}{\sqrt{l^2 - 1}} \]